Question: A secant line intersects the graph of $f(x)=x^2+5x$ at two points with $x$ -coordinates $3$ and $t$, where $t\neq 3$. What is the slope of the secant line in terms of $t$ ? Your answer must be fully expanded and simplified.
Answer: We are given that the secant line intersects the graph at $x=3$ and $x=t$. Since these points are on the the graph of $f(x)=x^2+5x$, we know that they must be $(3,24)$ and $(t,t^2+5t)$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{t^2+5t-24}{t-3} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{t^2+5t-24}{t-3}&=\dfrac{(t+8)(t-3)}{t-3} \\\\ &=t+8\text{, for }t\neq 3 \end{aligned}$ Since we are given that $t\neq 3$, we can conclude that the slope of the secant line is $t+8$.